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Mars2501 [29]
3 years ago
7

A sample of gas has a density of 0.53 g/L at 225 K and under a pressure of 108.8 kPa. Find the density of the gas at 345 K under

a pressure of 68.3 kPa. Assuming the mass is equal to 1 gram.
Chemistry
1 answer:
sukhopar [10]3 years ago
8 0

Answer:

\rho _2=0.22g/L

Explanation:

Hello!

In this case, since we are considering an gas, which can be considered as idea, we can write the ideal gas equation in order to write it in terms of density rather than moles and volume:

PV=nRT\\\\PV=\frac{m}{MM} RT\\\\P*MM=\frac{m}{V} RT\\\\P*MM=\rho RT

Whereas MM is the molar mass of the gas. Now, since we can identify the initial and final states, we can cancel out R and MM since they remain the same:

\frac{P_1*MM}{P_2*MM} =\frac{\rho _1RT_1}{\rho _2RT_2} \\\\\frac{P_1}{P_2} =\frac{\rho _1T_1}{\rho _2T_2}

It means we can compute the final density as shown below:

\rho _2=\frac{\rho _1T_1P_2}{P_1T_2}

Now, we plug in to obtain:

\rho _2=\frac{0.53g/L*225K*68.3kPa}{345K*108.8kPa}\\\\\rho _2=0.22g/L

Regards!

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Hydrazine (N 2 H 4 )) a rocket fuel reacts with oxygen to form nitrogen gas and water vapor . The reaction is represented with t
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The mass of hydrazine (N₂H₄) required to produce 96 g of water (H₂O) is 85.4 g (Option C)

<h3>Balanced equation </h3>

N₂H₄ + O₂ —> N₂ + 2H₂O

Molar mass of N₂H₄ = (2×14) + (4×1) = 32 g/mol

Mass of N₂H₄ from the balanced equation = 1 × 32 = 32 g

Molar mass of H₂O = (2×1) + 16 = 18 g/mol

Mass of H₂O from the balanced equation = 2 × 18 = 36 g

SUMMARY

From the balanced equation above,

36 g of H₂O were produced by 32 g of N₂H₄

<h3>How to determine the mass of N₂H₄</h3>

From the balanced equation above,

36 g of H₂O were produced by 32 g of N₂H₄

Therefore,

96 g of H₂O will be produced by = (96 × 32) / 36 = 85.4 g of N₂H₄

Thus, 85.4 g of N₂H₄ is needed for the reaction

Learn more about stoichiometry:

brainly.com/question/14735801

8 0
2 years ago
The thallium (present as Tl2SO4) in a 9.486-g pesticide sample was precipitated as thallium(I) iodide. Calculate the mass percen
tino4ka555 [31]

Answer: The mass percentage of Tl_2SO_4 is 5.86%

Explanation:

To calculate the mass percentage of Tl_2SO_4 in the sample it is necessary to know the mass of the solute (Tl_2SO_4 in this case), and the mass of the solution (pesticide sample, whose mass is explicit in the letter of the problem).

To calculate the mass of the solute, we must take the mass of the TlI precipitate.  We can establish a relation between the mass of TlI and Tl_2SO_4 using the stoichiometry of the compounds:

moles\ of\ TlI = \frac{0.1824 g}{331.27\frac{g}{mol} } = 5.51*10^{-4}\ mol.

Since for every mole of Tl in TlI there are two moles of Tl in Tl_2SO_4, we have:

moles\ of\ Tl_2SO_4 = 2 * moles\ of\ TlI = 1,102*10^{-3}\ mol

Using the molar mass of Tl_2SO_4 we have:

mass\ of\ Tl_2SO_4 = 1,102*10^{-3}\ mol * 504.83\ \frac{g}{mol}= 0.56\ g

Finally, we can use the mass percentage formula:

mass\ percentage = (\frac{solute\ mass}{solution\ mass} )*100 = (\frac{mass\ of\ Tl_2SO_4}{pesticide\ sample\ mass})*100 = (\frac{0.56g}{9.486g})*100 = 5.86\%

6 0
3 years ago
Small quantities of oxygen can be prepared in the laboratory by heating potassium chlorate, KClO 3 ( s ) . The equation for the
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Answer:

Mass of O_2 produced = 32 g

Explanation:

Calculation of the moles of KClO_3 as:-

Mass = 82.4 g

Molar mass of KClO_3 = 122.55 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{82.4\ g}{122.55\ g/mol}

Moles= 0.67237\ mol

From the reaction shown below:-

2KClO_3\rightarrow 2KCl+3O_2

2 moles of potassium chlorate on reaction forms 3 moles of oxygen gas

So,

0.67237 moles of potassium chlorate on reaction forms \frac{3}{2}\times  0.67237 moles of oxygen gas

Moles of oxygen gas = 1 mole

Molar mass of oxygen gas  = 32 g/mol

<u>Mass of O_2 produced = 32 g</u>

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